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1 year ago

photon strikes the surface of tungsten and an electron is emitted. what is the maximum possible speed of the electron?

Physics

1 answer:

zalisa [80]1 year ago

3 0

Answer: The maximum possible speed v is √2( hν - Ф ) / m

Explanation: You could be referring to the provided explanation, despite the fact that the question isn't comprehensive. When a photon collides with the surface of any metal, it transmits all of its energy to the electron in the atom. The collision causes the electron to travel with a certain amount of kinetic energy. This is referred to as the photoelectric effect. The maximum kinetic energy is calculated using Einstein's equation for the photoelectric effect:

K.E. = hν - Ф

½ mv² = hν - Ф

Hence the maximum possible speed is:

v = √2( hν - Ф ) / m

For more information on the photoelectric effect refer to this link: brainly.com/question/25027428

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____ is the detection of halo objects like brown dwarfs when their gravitational fields concentrate the light of distant sources

maksim [4K]

Answer:

Microlensing.

Explanation:

This techniques is called Microlensing.

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2 years ago

** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

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$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

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2 years ago

A. If an electron in a hydrogen atom has an energy of −6.06 × 10^−20 J, which Bohr orbit is it in?

Orlov [11]

Explanation:

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3 years ago

An isolated charged point particle produces an electric field with magnitude E at a point 2 m away. At a point 1 m from the part

guajiro [1.7K]

Explanation:

The electric field at a distance r from the charged particle is given by :

$E=\dfrac{kq}{r^2}$

k is electrostatic constant

if r = 2 m, electric field is given by :

$E_1=\dfrac{kq}{(2)^2}\\\\=\dfrac{kq}{4}\ .....(1)$

If r = 1 m, electric field is given by :

$E_2=\dfrac{kq}{r_2^2}\\\\=\dfrac{kq}{1}\ ....(2)$

Dividing equation (1) and (2) we get :

$\dfrac{E_1}{E_2}=\dfrac{\dfrac{kq}{4}}{kq}\\\\\dfrac{E_1}{E_2}=\dfrac{1}{4}\\\\E_2=4\times E_1$

So, at a point 1 m from the particle, the electric field is 4 times of the electric field at a point 2 m.

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2 years ago

I NEED HELP ASAP IM ON TIMER !!!!!!!!

White raven [17]

Answer:

Explanation:

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